Solved with mrdriller, hakN, GuvercinciHoca, sevket12 Since $x\ge y$ and $a+b\ge x+y$, all points lie on the shaded region. Now, divide this region into $3$ parts as shown in the figure. First of all, if $x>a$, then for all sufficiently large $n$, we have $a^n+b^n\leq 2a^n<x^n\leq x^n+y^n$ since $\lim_{n\to \infty} \left(\frac xa\right)^n=\infty$. Hence, none of the points in the area $3$ satisfies the condition. Clearly, all the points in the area $2$ satisfy the condition since $a\ge x$ and $b\ge y$ if $(x,y)$ lies on the area $2$. Finally, let’s show that any point lying on the area $1$ satisfy the condition. See that if $(x,y)$ lies on the area $1$, then $a-x\ge y-b\ge 0$. Hence, $$a^n-x^n=(a-x)\left(\sum_{i=0}^{n-1} a^ix^{n-1-i}\right)\ge (y-b)\left(\sum_{i=0}^{n-1} b^iy^{n-1-i}\right)=y^n-b^n$$Thus, $a^n+b^n\ge x^n+y^n$. So the desired area equals to $(\text{total area of } 1,2,3)-(\text{area of } 3)=\frac{(a+b)^2}4-\frac{b^2}2=\frac{a^2+2ab-b^2}4$.